Oscillations Part 2

Concept:The time period $T$ of a simple pendulum depends on its length $L$ and gravity $g$ as $T = 2\pi \sqrt{L/g}$.For fixed $g$, the time period is proportional to the square root of the length: $T \propto \sqrt{L}$.Explanation:The initial length $L_1 = 30 \text{ cm}$.Number of oscillations $N_1 = 20$ in total time $t = 10 \text{ s}$.So initial time period $T_1 = t / N_1 = 10/20 = 0.5 \text{ s}$.The final number of oscillations $N_2 = 40$ in the same time $t = 10 \text{ s}$.So final time period $T_2 = t / N_2 = 10/40 = 0.25 \text{ s}$.Since $L \propto T^2$, we have $L_2 / L_1 = (T_2 / T_1)^2$.Substitute values: $L_2 / 30 = (0.25 / 0.5)^2 = (1/2)^2 = 1/4$.Thus $L_2 = 30 \times (1/4) = 7.5 \text{ cm}$.Answer:The required length is $7.5$ cm, corresponding to option (B).